Posted by **Rae** on Monday, July 2, 2012 at 3:00pm.

A steel storage tank for propane gas is to be constructed in the shape of a right circular cylinder with a hempishere at each end. The cost per square foot of constructing the end pieces is twice that of contructing the cylindrical piece. If the desired capacity of the tank is 144 pi cubic feet, what dimensions will minimize the cost of construction?

- Calculus -
**Reiny**, Monday, July 2, 2012 at 3:32pm
The two hemispheres at the ends will make up one complete sphere

Let the radius of the sphere be r ft, and let the height of the cylindrical part be h

Volume = cylinder + sphere

πr^2 h + (4/3)πr^3 = 144π

3πr^2 h + 4πr^3 = 432π

3r^2 h + 4r^3 = 432

h = (432 - 4r^3)/(3r^2)

The cost of production is dependent on the surface area of material used

SA = 4π^2 + 2πrh

= 4πr^2 + 2π(432 - 4r^3)/(3r^2)

= 4πr^2 + 288π/r^2 - (8/3)πr

d(SA)/dr = 8πr - 576π/r^3 - 8π/3 = 0 for a min of SA

8r - 576/r^3 - 8/3 = 0

times 3r^3

24r^4 -1728 - 8r^3 = 0

3r^4 - r^3 - 216 = 0

At this point I "cheated" a bit and ran it through Wolfram

http://www.wolframalpha.com/input/?i=3r%5E4+-+r%5E3+-+216+%3D+0

for a real solution of r=3, r = appr-3 and 2 complex roots,

so the only usable answer is r = 3

the h = (432 - 108)/27 = 12

So the cylinder part should be 12 ft long and the radius of each of the semispheres should be 3 ft

- Calculus -
**Damon**, Monday, July 2, 2012 at 5:22pm
remember the sphere surface is twice as expensive as the cylinder which has no compound curvature so is easy to roll.

c = 2(4 pi r^2) + 1(2 pi r h)

c= 8 pi r^2 + 2 pi r h

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